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Homework Help: Laplace Transform of sin(t)/t help

  1. Jun 17, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]\mathcal{L}\left[\frac{\sin(t)}{t}\right]=?[/tex]


    2. Relevant equations
    [tex]\mathcal{L}\left[f(t)\right]=\int_0^\infty e^{-zt}f(t)dt[/tex]


    3. The attempt at a solution
    [tex]\int_0^\infty e^{-zt}\frac{\sin(t)}{t}dt=\int_0^\infty \frac{e^{-zt}}{t}\left(\frac{e^{it}-e^{-it}}{2i}\right )dt[/tex]

    [tex]\int_0^\infty f(t)dt=\lim_{T\rightarrow\infty}\int_0^T \frac{1}{t}\left(\frac{e^{t(i-z)}-e^{-t(i+z)}}{2i}\right )[/tex]

    And I end up having to calculate things like
    [tex]\int_0^T\frac{e^{\alpha t}}{t}dt[/tex] wich doesn't seem to be the best way to do this.

    Does any one know how to do this properly?

    Any help would be appreciated, thanks in advance!
     
  2. jcsd
  3. Jun 17, 2010 #2
    First, note that as z tends to infinity, the integral approaches 0 since the kernel [itex]e^{-zt}[/itex] tends to 0 in the limit.

    The relevant tool is to differentiate under the integral sign by treating the integral as a function of z. This is justified by Leibniz's rule or more advanced tools from Lebesgue integration (e.g. Lebesgue's dominated convergence theorem). In fact, one approach is to make the substitution for sin as you did, and then differentiate under the integral sign in a distributional sense, and work with fourier transforms. But simply differentiating under the integral sign with respect to the parameter z from the beginning also suffices.
     
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